A transverse wave on a string is given by the formula D(x,t)=0.20sin(3.6x−25.0t), where D and x are in m and t is in s.
1.At t=0.12s, what is the transverse displacement of the point on the string where x=0.50m?
2.At t=0.12s, what is the (transverse) velocity of the point on the string where x=0.50m? Indicate the direction of velocity with the sign of your answer.
A transverse wave on a string is given by the formula D(x,t)=0.20sin(3.6x−25.0t), where D and x...
A transverse wave on a string is given by the formula D(x,t) = 0.15 sin(4.0x - 43.0t). where D and z are in m and t is in s Part AAt t = 7.0 x 10 2s. what is the transverse displacement of the point on the string where z-0.45m? Express your answer using two significant figures. D(0.45m, 7.0 × 10-2S) = Part B At t = 7.0 × 10 2s. what is the (transverse) velocity of the point on the string where x =...
A transverse wave on a cord is given by D(x,t)=0.15sin(3.2x−46.0t), where D and x are in m and t is in s. a)At t=6.6×10−2s, what is the displacement of the point on the cord where x=0.56m? b)At t=6.6×10−2s, what is the velocity of the point on the cord where x=0.56m?
A transverse wave on a cord is given by D(x,t)=0.18sin(2.7x−61.0t), where Dand x are in m and t is in s. At t=4.9×10−2s, what is the displacement of the point on the cord where x=0.66m? At t=4.9×10−2s, what is the velocity of the point on the cord where x=0.66m?
A transverse wave on a cord is given by D(x,t)=0.16sin(3.2x−46.0t), where Dand x are in m and t is in s part a At t=6.6×10^−2s, what is the displacement of the point on the cord where x=0.56m? D(0.56m,6.6×10^−2s)= part b At t=6.6×10^−2s, what is the velocity of the point on the cord where x=0.56m? ∂D/∂t(0.56m,6.6×10^−2s)= =
Given:
The equation describing a transverse wave on a string is
y(x,t)=( 4.00 mm )sin[( 162 s^−1 )t−( 42.5 m^−1 )x].
λ = 0.148 m
f= 25.8 Hz
A= 4mm
v= 3.81 m/s
Find the transverse displacement of a point on the string when t-0.180 s and at a position 0.145 m. Submit Previous Answers Request Answer
A transverse wave on a string is modeled with the wave function y(x, t) (0.80 m)sin[(0.85 m)x (1.70 s)t 0.20]. (Indicate the direction with the signs of your answers.) (a) Find the wave velocity (in m/s). m/s (b) Find the position (in cm) in the y-direction, the velocity (in cm/s) perpendicular to the motior of the wave, and the acceleration (in cm/s2) perpendicular to the motion of the wave of a small segment of the string centered at x 0.40...
A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 87 m/s. At t=0, the string particle at x = has a transverse displacement of 4.2 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = is 17 m/s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the wave equation...
transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 93 m/s. At t= 0, the string particle at x = 0 has a transverse displacement of 4.1 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 19 m/s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the...
For a certain transverse standing wave on a long string, an antinode is at x -0 and an adjacent node is atx0.30 m. The displacement y(t) of the string particle at x0 is shown in the figure, where the scale of the y axis is set by ys = 4.4 cm, when t = 0.50 s, what is the displacement of the string particle at (a) x = 0.50 m and (b) x = 0.40 m ? what is the...
The equation of a transverse wave traveling along a very long string is given by y = 6.1 sin(0.018πx + 3.1πt), where x and y are expressed in centimeters and t is in seconds. Determine the following values. (a) the amplitude cm (b) the wavelength cm (c) the frequency Hz (d) the speed cm/s (e) the direction of propagation of the wave +x−x +y−y (f) the maximum transverse speed of a particle in the string cm/s (g) the transverse displacement at...